Tot ce se afla sub radical se face conditie de existenta mai mare sau egal decat 0
1 CE x+1 >/ 0 => x>/ -1 , x apatine [-1,infinit)
V(x+1)=x-1 ridici totul la a 2
V(x+1)² = (x-1)²
x+1 = x²-2x+1
x²-2x-x+1-1=0
x²-3x=0
x(x-3)=0
x1=0
x2= 3
2. CE x-1 >/0 , x>/ 1
2-x >/0, -x>/ -2 , x</ 2 => x apartine [1,2]
V(x-1)+V(2-x)=1 ridici totul la a 2
x-1+2V(x-1)(2-x)+2-x = 1
1+2V(x-1)(2-x)=1
2V(x-1)(2-x)=0
(x-1)(2-x)=0
x1=1, x2= 2
3. CE x+1 >/0 , x>/ -1 x apartine [-1,infinit)
x²-x-2=0, x²+x-2x-2=0 , x(x+1)-2(x+1)=0 , (x+1)(x-2)=0 x apartine [-1,2]
Din ambele conditii am rezultat ca x apartine [-1,infinit]
V(x+1)=V(x²-x-2) ridici la a 2 si astfel ai scapat de radical
x+1=x²-x-2
x²-x-x-2-1=0
x²-2x-3=0
x²+x-3x-3=0
x(x+1)-3(x+1)=0
(x+1)(x-3)=0
x1= -1
x2= 3
Bafta!
